1. **State the problem:** Solve the differential equation $$x^2 y'' + 3x y' + 5y = 8x$$ with initial conditions $$y(1) = 2$$ and $$y(e^{\pi/4}) = 2 \sinh(\pi/4)$$. 2. **Identify the type of equation:** This is a Cauchy-Euler (or equidimensional) differential equation of the form $$x^2 y'' + a x y' + b y = f(x)$$. 3. **Solve the homogeneous equation:** $$x^2 y'' + 3x y' + 5y = 0$$. Assume solution $$y = x^m$$, then $$y' = m x^{m-1}, \quad y'' = m(m-1) x^{m-2}$$. Substitute into homogeneous equation: $$x^2 (m(m-1) x^{m-2}) + 3x (m x^{m-1}) + 5 x^m = 0$$ which simplifies to $$m(m-1) x^m + 3m x^m + 5 x^m = 0$$ $$x^m [m(m-1) + 3m + 5] = 0$$ $$m^2 + 2m + 5 = 0$$. 4. **Solve characteristic equation:** $$m^2 + 2m + 5 = 0$$ Discriminant $$\Delta = 2^2 - 4 \cdot 1 \cdot 5 = 4 - 20 = -16 < 0$$. Roots are complex: $$m = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i$$. 5. **Write homogeneous solution:** $$y_h = x^{-1} [C_1 \cos(2 \ln x) + C_2 \sin(2 \ln x)]$$. 6. **Find particular solution:** Since right side is $$8x$$, try $$y_p = A x$$. Calculate derivatives: $$y_p' = A$$ $$y_p'' = 0$$. Substitute into original equation: $$x^2 (0) + 3x (A) + 5 (A x) = 8x$$ $$3 A x + 5 A x = 8 x$$ $$8 A x = 8 x$$ Divide both sides by $$8x$$ (for $$x \neq 0$$): $$A = 1$$. So particular solution is $$y_p = x$$. 7. **General solution:** $$y = y_h + y_p = x^{-1} [C_1 \cos(2 \ln x) + C_2 \sin(2 \ln x)] + x$$. 8. **Apply initial conditions:** At $$x=1$$: $$y(1) = 1^{-1} [C_1 \cos(0) + C_2 \sin(0)] + 1 = C_1 + 1 = 2$$ $$\Rightarrow C_1 = 1$$. At $$x = e^{\pi/4}$$: Calculate $$y(e^{\pi/4})$$: $$y = (e^{\pi/4})^{-1} [1 \cdot \cos(2 \ln e^{\pi/4}) + C_2 \sin(2 \ln e^{\pi/4})] + e^{\pi/4}$$ Since $$\ln e^{\pi/4} = \pi/4$$, $$y = e^{-\pi/4} [\cos(2 \cdot \pi/4) + C_2 \sin(2 \cdot \pi/4)] + e^{\pi/4}$$ Evaluate trigonometric functions: $$\cos(\pi/2) = 0, \quad \sin(\pi/2) = 1$$ So $$y = e^{-\pi/4} [0 + C_2 \cdot 1] + e^{\pi/4} = C_2 e^{-\pi/4} + e^{\pi/4}$$ Given $$y(e^{\pi/4}) = 2 \sinh(\pi/4)$$ Recall $$\sinh t = \frac{e^t - e^{-t}}{2}$$, so $$2 \sinh(\pi/4) = e^{\pi/4} - e^{-\pi/4}$$ Set equal: $$C_2 e^{-\pi/4} + e^{\pi/4} = e^{\pi/4} - e^{-\pi/4}$$ Subtract $$e^{\pi/4}$$ from both sides: $$C_2 e^{-\pi/4} = - e^{-\pi/4}$$ Divide both sides by $$e^{-\pi/4}$$: $$C_2 = -1$$. 9. **Final solution:** $$\boxed{y = x^{-1} [\cos(2 \ln x) - \sin(2 \ln x)] + x}$$